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Energy and Mass Analysis of a Control Volume

  • #1

Homework Statement


Consider a system in which R 134-A is moved through an evaporator. The evaporator removes energy from the the surroundings at a rate of 0.54 kW. The pressure = 120 kPa and x = 0.2 right before the fluid enters the evaporator. After exiting the evaporator, the pressure = 120 kPa and T = -20 C. What is the mass flow rate through the evaporator?


Homework Equations


(1) [tex]\sum[/tex]mass_in = [tex]\sum[/tex]mass_out

(2) [tex]\dot{q}[/tex] - [tex]\dot{w}[/tex] = [tex]\sum[/tex](([tex]\dot{m}[/tex]_out) * (h + [(V_2)^2/2] + gz)) - [tex]\sum[/tex](([tex]\dot{m}[/tex]_in) * (h + [(V_1)^2/2] + gz))

(3) [tex]\dot{E}[/tex] = [tex]\dot{m}[/tex]e
where e = u + ke + pe​

The Attempt at a Solution


Assuming steady flow, equation 1 is applicable, so the mass flow rate going into the evaporator is the same as the mass flow rate exiting the evaporator. However, I'm not exactly sure how to go about from there.
 

Answers and Replies

  • #2
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0
Next you ask yourself:

Was there any work done to the system?
Was there any energy added (subtracted) from the system?
Did the kinetic energy change or remain nearly constant?
Are gravitational effects needed to be accounted for?
Was there a change in temperature? (assumptions about specific enthalpy)

Ask yourself these questions and you will find many variables go to zero and some stay. The solution is simple.
 

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